What is the projection of #(8i + 12j + 14k)# onto # (2i + 3j – 7k)#?

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Answer 1 · 2018-01-29T13:18:21

The vector projection is #=-36/sqrt62<2, 3,-7>#

The vector projection of #vecb# onto #veca# is

#proj_(veca)vecb=(veca.vecb)/(||veca||)^2veca#

#veca=<2,3,-7>#

#vecb= <8, 12,14>#

The dot product is

#veca.vecb =<2,3,-7>. <8,12,14> #

# = (2)*(8)+(3) *(12)+(-7)*(14)=16+36-84=-36 #

The modulus of #veca# is

#=||veca||=||<2,3,-7>|| =sqrt((2)^2+(3)^2+(-7)^2)=sqrt(4+9+49)=sqrt62#

Therefore,

#proj_(veca)vecb=-36/sqrt62<2, 3,-7>#